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Please help thanks

 Mar 7, 2019
 #1
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a Area UNDER the velocity-time graph is the dosplacement of the trains.

   train B has the larger area...it has travelled the farthest

   train b area = 8 x 5.8  - 1/2(.8)(8) =

   Train a area(displacement) = 8 x 5.8 - 1/2 (4.8)(8) =

 

b determint the equations of the lines for  train a and train b....    then    train a   = train b +9  and solve for t

 Mar 7, 2019
 #2
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y=.6t+1  for train A

 

y=.1t +5    for train B      .6t+1 = .1x+5  + 9      t = 26 seconds

 Mar 7, 2019

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